LeetCode #697 — EASY

Degree of an Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Example 1:

Input: nums = [1,2,2,3,1]
Output: 2
Explanation: 
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.

Example 2:

Input: nums = [1,2,2,3,1,4,2]
Output: 6
Explanation: 
The degree is 3 because the element 2 is repeated 3 times.
So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.

Constraints:

nums.length will be between 1 and 50,000.
nums[i] will be an integer between 0 and 49,999.

Step 01

Brute Force Baseline

Problem summary: Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements. Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[1,2,2,3,1]

Example 2

[1,2,2,3,1,4,2]

Core Insight

What unlocks the optimal approach

Say 5 is the only element that occurs the most number of times - for example, nums = [1, 5, 2, 3, 5, 4, 5, 6]. What is the answer?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #697: Degree of an Array
class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        Map<Integer, Integer> left = new HashMap<>();
        Map<Integer, Integer> right = new HashMap<>();
        int degree = 0;
        for (int i = 0; i < nums.length; ++i) {
            int v = nums[i];
            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
            degree = Math.max(degree, cnt.get(v));
            if (!left.containsKey(v)) {
                left.put(v, i);
            }
            right.put(v, i);
        }
        int ans = 1000000;
        for (int v : nums) {
            if (cnt.get(v) == degree) {
                int t = right.get(v) - left.get(v) + 1;
                if (ans > t) {
                    ans = t;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #697: Degree of an Array
func findShortestSubArray(nums []int) int {
	cnt := map[int]int{}
	left := map[int]int{}
	right := map[int]int{}
	var degree int
	for i, v := range nums {
		cnt[v]++
		if degree < cnt[v] {
			degree = cnt[v]
		}
		if _, ok := left[v]; !ok {
			left[v] = i
		}
		right[v] = i
	}
	ans := 100000
	for v, c := range cnt {
		if c == degree {
			t := right[v] - left[v] + 1
			if ans > t {
				ans = t
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #697: Degree of an Array
class Solution:
    def findShortestSubArray(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        degree = cnt.most_common()[0][1]
        left, right = {}, {}
        for i, v in enumerate(nums):
            if v not in left:
                left[v] = i
            right[v] = i
        ans = inf
        for v in nums:
            if cnt[v] == degree:
                t = right[v] - left[v] + 1
                if ans > t:
                    ans = t
        return ans

// Accepted solution for LeetCode #697: Degree of an Array
struct Solution;

struct Degree {
    left: usize,
    right: usize,
    count: usize,
}

use std::collections::HashMap;
use std::usize;

impl Solution {
    fn find_shortest_sub_array(nums: Vec<i32>) -> i32 {
        let mut hm: HashMap<i32, Degree> = HashMap::new();
        let n = nums.len();
        let mut max_degree: usize = 0;
        for i in 0..n {
            let x = nums[i];
            let e = hm.entry(x).or_insert(Degree {
                left: i,
                right: i,
                count: 0,
            });
            e.left = usize::min(e.left, i);
            e.right = usize::max(e.right, i);
            e.count += 1;
            max_degree = usize::max(e.count, max_degree);
        }
        let mut min_width: usize = n;
        for d in hm.values() {
            if d.count == max_degree {
                min_width = usize::min(d.right - d.left + 1, min_width);
            }
        }
        min_width as i32
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 2, 3, 1];
    assert_eq!(Solution::find_shortest_sub_array(nums), 2);
    let nums = vec![1, 2, 2, 3, 1, 4, 2];
    assert_eq!(Solution::find_shortest_sub_array(nums), 6);
}

// Accepted solution for LeetCode #697: Degree of an Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #697: Degree of an Array
// class Solution {
//     public int findShortestSubArray(int[] nums) {
//         Map<Integer, Integer> cnt = new HashMap<>();
//         Map<Integer, Integer> left = new HashMap<>();
//         Map<Integer, Integer> right = new HashMap<>();
//         int degree = 0;
//         for (int i = 0; i < nums.length; ++i) {
//             int v = nums[i];
//             cnt.put(v, cnt.getOrDefault(v, 0) + 1);
//             degree = Math.max(degree, cnt.get(v));
//             if (!left.containsKey(v)) {
//                 left.put(v, i);
//             }
//             right.put(v, i);
//         }
//         int ans = 1000000;
//         for (int v : nums) {
//             if (cnt.get(v) == degree) {
//                 int t = right.get(v) - left.get(v) + 1;
//                 if (ans > t) {
//                     ans = t;
//                 }
//             }
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.